Question

3) (2n²+2on²+n) divide l0n2

Solve full sum and show !!

Answer 1

Step-by-step explanation:

For every n∈Zn∈Z you have three possible cases. Either n=3kn=3k or n=3k+1n=3k+1 or n=3k+2n=3k+2 (for some k∈Zk∈Z).

Let us consider each of these cases separately:

If n=3kn=3k, then n2=(3k)2=3(3k2)n2=(3k)2=3(3k2), which means 3∣n23∣n2.

If n=3k+1n=3k+1, then n2=(3k+1)2=9k2+6k+1=3(3k2+2k)+1n2=(3k+1)2=9k2+6k+1=3(3k2+2k)+1. This implies 3∤n23∤n2.

If n=3k+2n=3k+2, then n2=(3k+2)2=9k2+12k+4=3(3k2+4k+1)+1n2=(3k+2)2=9k2+12k+4=3(3k2+4k+1)+1. So we have again 3∤n23∤n2.

So we see that if 3∤n3∤n (i.e., in the last two cases), then 3∤n23∤n2. This is the same as saying that 3∣n23∣n2 implies 3∣n3∣n.

Notice that we have in fact proved also that if 3∤n3∤n, then the remainder o